codeforces 803c

Posted 2020-10-10

Maximal GCD

题意：给出n，k，求k个严格递增的数，使得k个数之和等于n，并且要求k个数的gcd最大

思路：k*(k+1)/2 * gcd <= n，gcd是n的因子，先找出n的所有因子，然后找到最大的一个满足条件的

AC代码：

#include "iostream"
#include "iomanip"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define step(x) fixed<< setprecision(x)<<
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ll long long
#define endl ("\n")
#define ft first
#define sd second
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const ll mod=1e9+7;
const ll INF = 1e18+1LL;
const int inf = 1e9+1e8;
const double PI=acos(-1.0);
const int N=1e5+100;

ll n,k,t,u,gcd,g[N];
int main(){
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>n>>k;
    if(k>500000){
        cout<<"-1";
        return 0;
    }
    u=k*(k+1)/2;
    for(ll i=1; i*i<=n; ++i){
        if(n%i==0){
            if(i*i==n) g[++t]=i;
            else{
                g[++t]=i, g[++t]=n/i;
            }
        }
    }
    for(int i=1; i<=t; ++i){
        if(u<=n/g[i] && g[i]>gcd){
            gcd=g[i];
        }
    }
    if(gcd==0) cout<<"-1";
    else{
        ll v=n/gcd, last=v-u+k;
        for(int i=1; i<k; ++i) cout<<i*gcd<<" ";
        cout<<last*gcd<<endl;
    }
    return 0;
}